What is the general solution for 1=sin(t)+sqrt(3)cos(t) ?

The general solution for 1=sin(t)+sqrt(3)cos(t) is t= pi/2+2pin,t=(11pi)/6+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Trigonometry >

1=sin(t)+sqrt(3)cos(t)

Solution

1 = sin (t) + 3 cos (t)

Solution

t = \frac{π}{2} + 2 πn, t = \frac{11 π}{6} + 2 πn

Degrees

t = 9 0^{\circ} + 36 0^{\circ} n, t = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

1 = sin (t) + 3 cos (t)

Subtract

3 cos (t)

from both sides

sin (t) = 1 - 3 cos (t)

Square both sides

sin^{2} (t) = (1 - 3 cos (t))^{2}

Subtract

(1 - 3 cos (t))^{2}

from both sides

sin^{2} (t) - 1 + 23 cos (t) - 3 cos^{2} (t) = 0

Rewrite using trig identities

- 1 + sin^{2} (t) - 3 cos^{2} (t) + 2 cos (t) 3

Use the Pythagorean identity:

1 = cos^{2} (x) + sin^{2} (x)

1 - sin^{2} (x) = cos^{2} (x)

= - 3 cos^{2} (t) + 23 cos (t) - cos^{2} (t)

Simplify

= - 4 cos^{2} (t) + 23 cos (t)

- 4 cos^{2} (t) + 2 cos (t) 3 = 0

Solve by substitution

- 4 cos^{2} (t) + 2 cos (t) 3 = 0

Let:

cos (t) = u

- 4 u^{2} + 2 u 3 = 0

- 4 u^{2} + 2 u 3 = 0 : u = 0, u = \frac{3}{2}

- 4 u^{2} + 2 u 3 = 0

Write in the standard form

a x^{2} + b x + c = 0

- 4 u^{2} + 23 u = 0

Solve with the quadratic formula

- 4 u^{2} + 23 u = 0

Quadratic Equation Formula:

For

a = - 4, b = 23, c = 0

u_{1, 2} = \frac{- 2 3 \pm ( 2 3 ) ^{2} - 4 ( - 4 ) \cdot 0}{2 ( - 4 )}

u_{1, 2} = \frac{- 2 3 \pm ( 2 3 ) ^{2} - 4 ( - 4 ) \cdot 0}{2 ( - 4 )}

(23)^{2} - 4 (- 4) \cdot 0 = 23

(23)^{2} - 4 (- 4) \cdot 0

Apply rule

- (- a) = a

= (23)^{2} + 4 \cdot 4 \cdot 0

(23)^{2} = 2^{2} \cdot 3

(23)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} (3)^{2}

(3)^{2} : 3

Apply radical rule:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 3^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 3

= 2^{2} \cdot 3

4 \cdot 4 \cdot 0 = 0

4 \cdot 4 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 2^{2} \cdot 3 + 0

2^{2} \cdot 3 + 0 = 2^{2} \cdot 3

= 2^{2} \cdot 3

Apply radical rule:

n ab = n a n b,

assuming

a \geq 0, b \geq 0

= 3 2^{2}

Apply radical rule:

n a^{n} = a,

assuming

a \geq 0

2^{2} = 2

= 23

u_{1, 2} = \frac{- 2 3 \pm 2 3}{2 ( - 4 )}

Separate the solutions

u_{1} = \frac{- 2 3 + 2 3}{2 ( - 4 )}, u_{2} = \frac{- 2 3 - 2 3}{2 ( - 4 )}

u = \frac{- 2 3 + 2 3}{2 ( - 4 )} : 0

\frac{- 2 3 + 2 3}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 2 3 + 2 3}{- 2 \cdot 4}

Add similar elements:

- 23 + 23 = 0

= \frac{0}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{0}{- 8}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{0}{8}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

u = \frac{- 2 3 - 2 3}{2 ( - 4 )} : \frac{3}{2}

\frac{- 2 3 - 2 3}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 2 3 - 2 3}{- 2 \cdot 4}

Add similar elements:

- 23 - 23 = - 43

= \frac{- 4 3}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 4 3}{- 8}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{4 3}{8}

Cancel the common factor:

4

= \frac{3}{2}

The solutions to the quadratic equation are:

u = 0, u = \frac{3}{2}

Substitute back

u = cos (t)

cos (t) = 0, cos (t) = \frac{3}{2}

cos (t) = 0, cos (t) = \frac{3}{2}

cos (t) = 0 : t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

cos (t) = 0

General solutions for

cos (t) = 0

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

cos (t) = \frac{3}{2} : t = \frac{π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn

cos (t) = \frac{3}{2}

General solutions for

cos (t) = \frac{3}{2}

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

t = \frac{π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn

t = \frac{π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn

Combine all the solutions

t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn, t = \frac{π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

sin (t) + 3 cos (t) = 1

Remove the ones that don't agree with the equation.

Check the solution

\frac{π}{2} + 2 πn :

True

\frac{π}{2} + 2 πn

Plug in

n = 1

\frac{π}{2} + 2 π 1

For

sin (t) + 3 cos (t) = 1

plug in

t = \frac{π}{2} + 2 π 1

sin (\frac{π}{2} + 2 π 1) + 3 cos (\frac{π}{2} + 2 π 1) = 1

Refine

1 = 1

\Rightarrow True

Check the solution

\frac{3 π}{2} + 2 πn :

False

\frac{3 π}{2} + 2 πn

Plug in

n = 1

\frac{3 π}{2} + 2 π 1

For

sin (t) + 3 cos (t) = 1

plug in

t = \frac{3 π}{2} + 2 π 1

sin (\frac{3 π}{2} + 2 π 1) + 3 cos (\frac{3 π}{2} + 2 π 1) = 1

Refine

- 1 = 1

\Rightarrow False

Check the solution

\frac{π}{6} + 2 πn :

False

\frac{π}{6} + 2 πn

Plug in

n = 1

\frac{π}{6} + 2 π 1

For

sin (t) + 3 cos (t) = 1

plug in

t = \frac{π}{6} + 2 π 1

sin (\frac{π}{6} + 2 π 1) + 3 cos (\frac{π}{6} + 2 π 1) = 1

Refine

2 = 1

\Rightarrow False

Check the solution

\frac{11 π}{6} + 2 πn :

True

\frac{11 π}{6} + 2 πn

Plug in

n = 1

\frac{11 π}{6} + 2 π 1

For

sin (t) + 3 cos (t) = 1

plug in

t = \frac{11 π}{6} + 2 π 1

sin (\frac{11 π}{6} + 2 π 1) + 3 cos (\frac{11 π}{6} + 2 π 1) = 1

Refine

1 = 1

\Rightarrow True

t = \frac{π}{2} + 2 πn, t = \frac{11 π}{6} + 2 πn

Graph

Popular Examples

cosh(x)= 5/4

sec(θ)-sqrt(2)tan(θ)=0

2sin(x-pi/3)=-sqrt(2)

3sin(θ)=1+cos(θ)

sqrt(2)sin(x)-sqrt(2)cos(x)=2

Frequently Asked Questions (FAQ)

What is the general solution for 1=sin(t)+sqrt(3)cos(t) ?
The general solution for 1=sin(t)+sqrt(3)cos(t) is t= pi/2+2pin,t=(11pi)/6+2pin